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September 11, 2013, 01:19 |
Boundary Condition Problem
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#1 |
New Member
Shu Shu Chang
Join Date: Sep 2013
Location: Taiwan, Tainan
Posts: 2
Rep Power: 0 |
Hi ANSYS users,
I am a new FLUENT user, this is my first post in CFD online, maybe this is a easy question. there is a channel have a inlet and a outlet, the inlet velocity is a sine function depend on time, and the outlet just outflow. As you know, the sine wave is positive at 0 to PI, negative at PI to 2*PI. When time at PI to 2*PI, the velocity become the outlet velocity but reverse the direction, and the inlet become outflow. Easy to say, the inlet and outlet boundary condition exchange. I have a problem now. How to set the boundary condition as the example in fluent? Should I use UDFs? BTW, I had tried using "Dynamic mesh -> event", but appear the error message "the BCs are same type", maybe it's a wrong way. Please give me some suggest. Thanks |
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September 11, 2013, 07:51 |
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#2 |
New Member
Andrew Walmsley
Join Date: Sep 2013
Posts: 5
Rep Power: 12 |
With regards to the transient boundary condition, yes you will to use UDFs. There are quite a few examples around online for transient boundary conditions. Alas, none for what I need.
I can't help with the other issue, sorry I don't have experience with dynamic meshing as of this time. |
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September 11, 2013, 21:54 |
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#3 |
New Member
Join Date: Sep 2013
Location: Shanghai China
Posts: 6
Rep Power: 12 |
Maybe you can use velocity inlet and velocity outlet, or velocity inlet and pressure outlet.
Is the flow fluid compressible or not? The compressibility will affect this periodic flow process. |
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September 12, 2013, 01:32 |
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#4 |
New Member
Shu Shu Chang
Join Date: Sep 2013
Location: Taiwan, Tainan
Posts: 2
Rep Power: 0 |
Thank quantities, AWalmsley11 reply.
Maybe I had found the way to solve this problem. I let FLUENT calculate the time at 0 to PI, when time equal to PI, FLUENT calculate completed. And I can change the boundary condition between inlet and outlet, and don't initialized solution, let FLUENT keep calculating time at PI to 2*PI. This way can work, but I don't know it will cause what bad effect. |
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September 12, 2013, 09:38 |
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#5 |
New Member
Andrew Walmsley
Join Date: Sep 2013
Posts: 5
Rep Power: 12 |
If it's done that way it will not have the sine function, it will be constant for that time. Think of it as it will appear as a digital signal rather than analogue in graphical terms.
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